Q118 / 180 · KCET 2011 · Mathematics
In Argand's plane, the point corresponding to $\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}$ lies in
- Aquadrant I
- Bquadrant II
- Cquadrant III
- Correct answerDquadrant IV
Solution
Given,
$$
\begin{aligned}
&\frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}=\frac{(1-i \sqrt{3}+i+\sqrt{3})(\sqrt{3}-i)}{(3+1)} \\
&=\frac{1}{4} \cdot\{(1+\sqrt{3})+i(1-\sqrt{3})\} \cdot(\sqrt{3}-i) \\
&=\frac{1}{4} \cdot\{\sqrt{3}(1+\sqrt{3})+i(1-\sqrt{3}) \sqrt{3} \\
&\left.=\frac{1}{4} \cdot\{\sqrt{3}+3+1-\sqrt{3})+(\sqrt{3}-3-1-\sqrt{3}) i\right\} \\
&=\frac{1}{4} \cdot\{4-4 i\}=1-i
\end{aligned}
$$
The point $(1-i)$ in Arg and plane is $(1,-1)$ which lies in IVth quadrant.
Hence, the correct answer is option: 'D'.